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I am trying to find the gain (Vo/Vi) of the following inverting amplifier circuit. enter image description here

In a normal inverting amplifier circuit with input resistor \$R1\$, I can simply use $$ \frac{Vo}{Vi} = \frac{-Rf}{R1} $$ to get the result. However, in this circuit, the input resister R1 is removed.

Assuming it is an ideal amplifier, no current goes in to negative terminal. So, current \$I1\$ and \$I2\$ are the same.

$$ I2 = \frac{(Vo - 0)}{100k} $$

I am not sure how I can include \$ Vi \$ in the equation so that I can calculate the gain.

If I use \$ Vo = A (V^{-}- V^{+}) \$ , since \$ V^{+} = V^{-} = 0 \$, wouldn't it just give me 0?

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    \$\begingroup\$ If the voltage \$V_i\$ is set by a voltage source then how can the 100 kohm resistor do anything at all? You need to think about the fact if this circuit is an amplifier that needs a voltage applied to the input or that it needs a current applied to the input. \$\endgroup\$ Dec 22 '20 at 14:52
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What you have drawn is a transimpedance amplifier; input is current (and not voltage because it has a virtual ground at -Vin) and the output is voltage: -

enter image description here

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But with those values you can't calculate voltage gain.

R1 would be 0 ohms here, so based on that, voltage gain would be infinite. Or rather, not defined due to dividing by zero.

Other way to lool at it is that for an ideal op-amp, Vin must equal to 0V because inverting and non-inverting input must have equal voltages. Since feeding any input voltage other than zero will violate this rule, the output will just saturate, i.e. the gain is infinite again.

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The circuit in your question is a transimpedance amplifier. The output voltage is proportional to the input current, as you have surmised.

\$v_{out} = -i_{in}R_f\$

The amplifier will attempt to keep the input voltage \$v_{in}\$ near zero.

(Unrelated to your question, the amplifier may require a small capacitor in parallel to the feedback resistor for stabilitly.)

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